How to find $\mathbb{P}\left(\int_0^1W_tdt>\dfrac{2}{\sqrt{3}}\right)$?

Question

How to find $\mathbb{P}\left(\int_0^1W_tdt>\dfrac{2}{\sqrt{3}}\right)$? Where $W_t$ is Browninan Motion.

What I have done is, for $Z\sim N(0,1)$:

$$\begin{align} \mathbb{P}\left(\int_0^1W_tdt>\dfrac{2}{\sqrt{3}}\right) &= \mathbb{P}\left(\int_0^1\sqrt{t}Zdt>\dfrac{2}{\sqrt{3}}\right)\\ &= \mathbb{P}\left(Z\dfrac{2}{3}>\dfrac{2}{\sqrt{3}}\right)\\ &= \mathbb{P}\left(Z>\sqrt{3}\right)\\ &= \dfrac{1}{\sqrt{2\pi}}\int_\sqrt{3}^\infty e^{-\frac{z^2}{2}}dz\\ &= -N\left(-\sqrt{3}\right) \end{align}$$

Any help would be very much appreciated.

Answer 1

We first show that $$ \int_0^t W_s \,\mathrm{d}s \sim N\left(0,\frac13 t^3 \right).$$ Indeed, the integral is defined as a limit of Riemann sums; since each of these sums are normally distributed, the limiting integral is also normally distributed (this can be proven using characteristic functions, for instance). Applying Fubini's theorem, $$ \mathbb{E} \left[\int_0^t W_s \,\mathrm{d}s \right] = \int_0^t \mathbb{E}[W_s] \,\mathrm{d}s = \int_0^t 0 \,\mathrm{d}s = 0,$$ where the interchange of limits is permissible since w.p.1 the function $s \mapsto W_s$ is continuous and hence bounded on a compact interval. Finally, \begin{align*} \mathrm{Var}\left(\int_0^t W_s \,\mathrm{d}s\right) &= \mathbb{E}\left[\left(\int_0^t W_s \,\mathrm{d}s\right)^2\right] \\ &= \mathbb{E}\left[\left(\int_0^t W_s \,\mathrm{d}s\right)\left(\int_0^t W_u \,\mathrm{d}u\right)\right] \\ &= \mathbb{E}\left[\int_0^t \int_0^t W_sW_u \,\mathrm{d}s \mathrm{d}u \right] \\ &= \int_0^t \int_0^t \mathbb{E}[W_sW_u] \,\mathrm{d}s \mathrm{d}u \\ &= \int_0^t \int_0^t \min(s,u) \,\mathrm{d}s \mathrm{d}u \\ &= 2 \int_0^t \int_0^u s \,\mathrm{d}s \mathrm{d}u \\ &= \int_0^t u^2 \mathrm{d}u \\ &= \frac{1}{3}t^3, \end{align*} where we use the fact that $\mathrm{Cov}(W_s, W_t) = \min(s, t)$ and Fubini again to pass the expectation inside. It follows that $$ \sqrt{3} \int_0^1 W_s \,\mathrm{d}s \sim N(0,1), $$ and $$\mathbb{P}\left(\int_0^1 W_s \,\mathrm{d}s \geq \frac{2}{\sqrt{3}} \right) = 1 - \Phi(2),$$ where $\Phi$ is the standard normal cdf.

Answer 2

The law of $(W_t)_{0\leq t\leq 1}$ is not the same as that of $(\sqrt{t}Z)_{0\leq t \leq 1}$. So there is no reason $\int_{0}^{1} W_t \, dt$ and $\int_{0}^{1} \sqrt{t}Z \, dt$ have the same law, and as it turns out they have different laws.

Since $X = \int_{0}^{1} W_t \, dt$ is gaussian, it suffices to determine its mean and variance. And they can be computed as

$$ \mathbb{E}[X]=\int_{0}^{1} \mathbb{E}[W_t] \, dt = 0$$

and

$$ \mathbb{E}[X^2] = \int_{0}^{1} \int_{0}^{1} \mathbb{E}[W_s W_t] \, dsdt = \int_{0}^{1} \int_{0}^{1} s \wedge t \, dsdt = \frac{1}{3}. $$

So $X \sim \mathcal{N}(0, \frac{1}{3})$ and hence $X \stackrel{d}{=} \frac{1}{\sqrt{3}}Z$. This gives

$$ \mathbb{P}(X > \tfrac{2}{\sqrt{3}}) = \mathbb{P}(Z > 2). $$